If $\frac a 3>1$ then the local maximum doesn't lie in the range $-\frac 1 3 \le x \le 1$. The maximal value will be at one of the end points.
So if $\frac a 3 > 1 \implies a>3$ then the maximum value will be at $f(1)=3a-6$. $\frac{a^3}9$ will be a bigger value but $x=\frac a 3$ will not lie in the range of $x$ when $a>3$.